Let $a(x)=x^6+5$, and $b(x)=x^3+x^2$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{x^6+5}{x^3+x^2}$ : First, we divide ${x^3}$ into ${x^6}$ and get ${x^3}$ : $ \hphantom{1567|1.4} {x^3}\\ {{{x^3}+x^2}}|\overline{{x^6}+0x^5+5}\\ \hphantom{37.......|}\llap{-}\underline{(x^6+x^5)}\\ \hphantom{37|3..........}-x^5\\ $ [What did we do here?] Next, we divide ${x^3}$ into ${-x^5}$ to get ${-x^2}$, and continue doing this until we find the quotient: $ \hphantom{1567|14} {x^3 \ {- \ \ x^2}\ +\ \ x \ - \ 1}\\ {{{x^3}+x^2}}|\overline{x^6+0x^5+0x^4+0x^3+0x^2+5}\\ \hphantom{37.......|}\llap{-}\underline{(x^6+x^5)}\\ \hphantom{37|3...........}{-x^5}+0x^4\\ \hphantom{37.............|}\llap{-}\underline{(-x^5-x^4)}\\ \hphantom{37|3.......................}{+x^4+0x^3}\\ \hphantom{3788888888888.......|}\llap{-}\underline{(x^4+x^3)}\\ \hphantom{37|3...............99999........}{-x^3+0x^2}\\ \hphantom{3788888888888.............|}\llap{-}\underline{(-x^3-x^2)}\\ \hphantom{37|3...............9999999999..........}{+x^2+5}\\ $ [What did we do here?] The process stops here because $x^3+x^2$ is a polynomial of the third degree and $x^2+5$ is a polynomial of the second degree. So it follows that ${r(x)}={x^2+5}$, ${q(x)}={x^3-x^2+x-1}$, and $ \dfrac{x^6+5}{x^3+x^2}={x^3-x^2+x-1}+\dfrac{{x^2+5}}{x^3+x^2}$ To conclude, $q(x)=x^3-x^2+x-1$ $r(x)=x^2+5$